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1 Example 2 Estimate by the six Rectangle Rules using the regular partition P of the interval [0, ] into 6 subintervals. Solution Observe that the function with g(0)=1 is continuous at x=0 because Note that P = {0, /6, /3, /2, 2 /3, 5 /6, } with each subinterval of width /6. The six subintervals are: [0, /6 ], [ /6, /3], [ /3, /2], [ /2, 2 /3], [2 /3, 5 /6] and [5 /6, ]. Then The L k are the heights of the rectangles used to approximate this definite integral. In the Left Endpoint Rule the L k are the values of g on the left endpoints of the six subintervals: g(0), g( /6), g( /3), g( /2), g(2 /3), g(5 /6).

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2 In the Right Endpoint Rule the L k are the values of g on the right endpoints of the six subintervals: g( /6), g( /3), g( /2), g(2 /3), g(5 /6), g( ). In the Midpoint Rule the L k are the values of g on the midpoints of the six subintervals: g( /12), g( /4), g(5 /12), g(7 /12), g(3 /4), g(11 /12). In the Trapezoid Rule the L k are the averages of the values of g on the endpoints of each of the six subintervals: ½[g(0)+ g( /6)], ½[g( /6)+ g( /3)], ½[g( /3)+ g( /2)], ½[g( /2)+ g(2 /3)], ½[g(2 /3)+ g(5 /6)], ½[g(5 /6)+ g( )]. Therefore, the Lower Riemann sum coincides with the estimate of the Right Endpoint Rule and the Upper Riemann sum coincides with estimate of the Left Endpoint Rule. Since the function g is decreasing on [0,1], it has its maximum value at the left endpoint of each subinterval and its minimum value at the right endpoint of each subinterval. The values of the L k are summarized in the table on the next slide The six subintervals are: [0, /6 ], [ /6, /3], [ /3, /2], [ /2, 2 /3], [2 /3,5 /6], [5 /6, ].

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3 The Left Endpoint Rule and Upper Riemann Sum give the same estimate: The Right Endpoint Rule and Lower Riemann Sum give the same estimate: The Midpoint Rule gives the estimate: The Trapezoid Rule gives the estimate: Left Endpoint Rule Right Endpoint Rule Midpoint Rule Trapezoid Rule Lower Riemann Sum Upper Riemann Sum L1L2L3L4L5L6L1L2L3L4L5L6 g(0)=1 g( /6) .955 g( /3) .827 g( /2) .637 g(2 /3) .414 g(5 /6) .191 g( /6) .955 g( /3) .827 g( /2) .637 g(2 /3) .414 g(5 /6) .191 g( )=0 g( /12) .989 g( /4) .900 g(5 /12) .738 g(7 /12) .527 g(3 /4) .300 g(11 /12) .090 ½(1+.955) ½(.955+.827) ½(.827+.637) ½(.637+.414) ½(.414+.191) ½(.191+0) g( /6) .955 g( /3) .827 g( /2) .637 g(2 /3) .414 g(5 /6) .191 g( )=0 g(0)=1 g( /6) .955 g( /3) .827 g( /2) .637 g(2 /3) .414 g(5 /6) .191

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